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ANSYS ELEMENT BEAM188

ANSYS has replaced the beam element (BEAM3) based on the elementary beam theory by the element BEAM188 based on the Timoshenko beam theory. In order to use their improved element, one has to specify certain cross section parameters and certain options for the element. I aim to compare here the implementation by ANSYS with a direct solution of the Timoshenko equations in order to shed some light on what parameters and options to use and the accuracy of the ANSYS model. 1. Timoshenko Beam Theory 1.1 General equations for bending in a plane The strain energy in shear is neglected in the elementary beam model. Timoshenko modified the elementary theory to account for additional deformations due to shear by introducing a shear correction factor. The resulting approximate equations for bending of a symmetrical section in the x-y plane are as follows (Ref. 3).

Fig. 1.1 Sign Convention

Equilibrium equations:

( ) 0,

( ) ( ) 0.

dV x pdxdM x V xdx

+ =

+ = (1.1)

Geometry of deformation:

( , ) ( ) ,( , ) ( ),

,( ) ( ),

( ) .

u x y x yv x y v x

ydv x xdxd xdx

θ

ε κ

γ θ

θκ

= −=

= −

= −

=

(1.2)

Elasticity:

,.

EG

σ ετ γ

==

(1.3)

Stress Resultants:

2

,

.A

A

M y dA EI

V dA kGA

σ κ

τ γ

= − =

= =

∫∫

(1.4)

The shear factor k is introduced to account for the nonlinear distribution of shear strain over the cross section. I will address the calculation of the shear factor in section 1.6. 1.2. Basic equations to be solved: The resulting differential equations for the stress resultants and bending deformations are as follows.

( ) ( ) 0,

( ) ( ) 0,

( ) ( ),

( ) ( ) ( ).

dV x p xdxdM x V xdxd xEI M xdxdv x EIEI EI x V xdx kGA

θ

θ

+ =

+ =

=

= +

(1.5)

1.3. General solution when the loads are only at the ends of the beam For the case ( ) 0p x = , the general solution to the differential equations (1.5) is:

1

1 2

2

1 2 3

3 2

1 2 3 4

( ) ,

( ) ,

( ) ,2

( ) .6 2

V x cEIM x c x cEI

xx c c x c

EI x xv x c x c c x ckGA

θ

=

= − +

= − + +

⎛ ⎞= − + + +⎜ ⎟

⎝ ⎠

(1.6)

The following sign convention for end displacement and rotation is used here:

Fig. 1.2 Sign convention for end displacements

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The constants are therefore:

( ) ( ) ( )

( ) ( )

4 1

3 1

2 2 1 2 1 2 12

1 2 1 2 13 2

,,1 3 6 ,

12 6 ,

c vc

c v vL L L

c v vL L

θβ βθ θ θ θ

β β θ θ

==

= − − + + −

= − − +

(1.7)

where

1

2

121 EIkAGL

β−

⎛ ⎞= +⎜ ⎟⎝ ⎠ . (1.8)

The following sign convention for end force and moment is used here:

Fig. 1.3 Sign convention for end forces

Using the solution (1.6) and (1.7), =F KD (1.9) where

1 1

1 1

2 2

2 2

,,f vMf vM

θ

θ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

F D

(1.10)

and the stiffness matrix K is

( ) ( )

( ) ( )

3 2 3 2

2 2

3 2 3 2

2 2

12 6 12 6

6 63 1 3 1.

12 6 12 6

6 63 1 3 1

EI EI EI EIL L L LEI EI EI EIL L L LEI EI EI EIL L L LEI EI EI EIL L L L

β β β β

β ββ β

β β β β

β ββ β

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥+ − −⎢ ⎥= ⎢ ⎥−− −⎢ ⎥⎢ ⎥⎢ ⎥−− +⎢ ⎥⎣ ⎦

K

(1.11)

If the shear deformation is neglected ( 2 /AL I →∞ ), then 1β = and the stiffness matrix reduces to that of the elementary beam theory [Ref. 5, eq. 1.82].

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1.4 Virtual work and potential energy From the equilibrium equations (1.1),

( ) ( ) ( ) 0dV x dM xp v V x dxdx dx

δ δθ⎧ ⎫⎛ ⎞ ⎛ ⎞+ + + =⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎩ ⎭∫ (1.12)

For all functions (variations) ( )v xδ and ( )xδθ . Integrating by parts gives the (virtual work) rquation: ( ) [ ] [ ].M V dx p vdx V v Mδκ δγ δ δ δθ+ = + +∫ ∫ (1.13) The nodal forces for the distributed loading p(x) are therefore defined by T ( ) ( ) .p x v x dx= ∫F D . (1.14) For the case p is constant, using (1.6), we find

1 1 2 2

2 2T

2 12 2 12

v v

pL pL pL pL

θ θ⎡ ⎤

= −⎢ ⎥⎣ ⎦

F (1.15)

Inclusion of the shearing deformation leads to a coupling of the force and moment. 1.4. Cantilever beam with end load.

Fig. 1.4 End loaded beam

Applying (1.9) with 1 1 2 20, 0, 0, ,v M f Pθ = = = = leads to

2

2

3

2 2

,2

31 .3

PLEIPL EIvEI kGAL

θ =

⎛ ⎞= +⎜ ⎟⎝ ⎠

(1.16)

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For a square cross section, 2, 2, 4, 4 / 3b h A I= = = = , with 20L = , 1000E = , 0.3ν = , and using 0.833333k = ,

3

3 ( ) 1.0078EIv LPL

= . (1.17)

The inclusion of the shear correction increases the displacement by 0.78% in this case. 1.5. Cantilever with a uniformly distributed load of intensity p (constant).

Applying (1.9) with 2

1 1 2 20, 0, , ,12 2pL pLv M fθ = = = − = as dictated by (1.15), leads to

3

2

4

2 2

,6

41 .8

pLEIpL EIvEI kGAL

θ =

⎛ ⎞= +⎜ ⎟⎝ ⎠

(1.18)

For the case 1, 1000, 0.3, 4 / 3, 2, 2, 4, 20, 0.833333,p E I b h A L kν= = = = = = = = =

24

8 1.0104EI vpL

= (1.19)

The inclusion of the shear correction increases the displacement by 1.04% in this case. 1.6. Value of the shear factor k I regard the bean theory as an approximation to the three dimensional equations of linear elasticity. The value of k should be that one which gives the most accurate approximation to the solution of the 3D equations. Unfortunately, that value depends on the three dimensional problem that we want to approximate and, in particular, whether the problem is one of equilibrium, vibration, or wave propagation. There is therefore no universally best value for k. Pilkey (Ref. 3, section 6.3) provides a summary of various methods that have been proposed for the calculation of k. I will consider here the result if the shear factor is determined by comparing the strain energy in the statical problem for the Timoshenko model with other analytical solutions.

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1.6.1 Circular Cross Section

Fig. 1.5 Circular cross section

For a circular cross section of radius R, the 3D solution for shear stress in bending by the end load P is (Ref. 4):

( )2 2 2 ,

,xy

xz

c R y az

byz

τ

τ

= − −

= (1.20)

where

41 2 1 2 3 2, , , .

3 2 4(1 ) 8(1 ) 4P P Ra b c II I

ν ν ν πν ν ν

− + += = = =+ + +

(1.21)

The shear strain energy per unit length is

( )

( )

22 2

2

7 14 81 1 .2 2 6 1

xy xz PU dAG G G A

ν ντ τν

+ +⎛ ⎞= + =⎜ ⎟⎜ ⎟ +⎝ ⎠

∫∫ (1.22)

For the Timoshenko model,

21

2 2PU VGAk

γ= = (1.23)

Equating these two expressions for the energy gives

( )22

6 1.

7 14 8k

νν ν+

=+ +

(1.24)

Note that k depends on the Poisson ratio and 6 / 7 0.8571k = = for 0ν = .

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1.6.2 Rectangular Cross Section Determination of the shear factor is more complicated for the end loaded beam with a rectangular cross section because there is no known simple solution.

Fig. 1.6 Rectangular Cross Section

(a) Elementary beam model From the elementary beam theory, the shearing strain energy per unit length is

2

2xyU dAG

τ= ∫ . (1.25)

For the Timoshenko model,

.21

2 2VU VGAk

γ= = (1.26)

Equating the two expressions gives

2

2xy

VkA dAτ

=∫

(1.27)

For the rectangle, using the stress from the elementary beam model,

2

2

3 212xy

VQ V yIb A h

τ ⎛ ⎞= = −⎜ ⎟

⎝ ⎠ , (1.28)

gives

56

k = . (1.29)

Note that the shear component xzτ is neglected in the elementary beam theory.

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(b) Compare with an exact plane stress solution The following plane stress problem is a model of the end loaded cantilever beam.

Fig. 1.7 Plane stress model

The top ( y c= ) and bottom ( y c= − ) edges are free of load. The right end ( 0x = ) has a shearing load of magnitude P that is distributed parabolically. The left end ( x L= ) has a parabolic shear load and a linearly varying normal traction:

2

2

2

2 3

30 : 1 .4

3 3: 1 , .4 2

xy

xy x

P yxc c

P y PLyx Lc c c

τ

τ σ

⎛ ⎞= = −⎜ ⎟

⎝ ⎠⎛ ⎞

= = − =⎜ ⎟⎝ ⎠

(1.1.30)

The clamped end is modeled by ( ,0) 0u L = , ( ,0) 0v L = , ( , ) 0u L c± = , The exact solution for plane stress is (Ref. 5, pg. 38):

( )2

2

3, 14

xy V yx yG cG cτ

γ ⎛ ⎞= = −⎜ ⎟

⎝ ⎠ . (1.31)

The shearing strain energy is

2

0

1 32 5

c L

xy xyc

V LU dydxAG

τ γ+

−= =∫ ∫ (1.32)

The cantilever beam modeled by the Timoshenko equations, section 4, has a shear strain energy

21 ( )

2 2V LU Vv LkAG

= = (1.33)

Equating the shear strain energy gives

56

k = (1.34)

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(e) Three Dimensional Elasticity The exact three dimensional equations for stress in the beam were solved in Ref. 1 and Ref. 3 to determine the strain energy. For each cross section, the differential equations were solved by the finite element method using the nine node Lagrangean element. The results depend on the value of the Poisson ratio. For the case / 2h b = , the result (Ref. 3, pg. 265) is

0 : 0.8333,0.3 : 0.8329,0.5 : 0.8325.

kkk

ννν

= == == =

(1.35)

The value depends on the Poisson ratio but differs little from 5 / 6 for h and b of the same order of magnitude. 2. ANSYS Element BEAM 188 According to the Element Reference manual:

This element is based on the Timoshenko beam theory. It is a two–node 3D element with seven degrees of freedom at each end: translations, rotations, and warping. There are three models selected by the key option three. When KEYOPT (3) = 0 (linear, default), linear shape functions and one-point integration are used; therefore, all element solution quantities are constant. When KEYOPT (3) = 2 (quadratic), the element has an internal node. The shape functions are effectively quadratic. Two-point integration is used, resulting in linear variation of element solution quantities. Linearly varying bending moments are represented exactly. When KEYOPT (3) = 3 (cubic), The element has two internal nodes and cubic shape functions with three-point integration, resulting in quadratic variation of solution quantities. Quadratically varying bending moments are represented exactly. Distributed normal loads can be applied to the element by the SFBEAM command.

We will examine the results of using this element for some examples to see what actually results, without looking into the derivation of the stiffness matrix and loading parameters, for bending in a plane without twist.

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2.1 ANSYS value for the shear factor According to the manual,

the shear-correction factor is calculated in accordance with Ref. 1 and Ref. 2. Each section is assumed to be an assembly of predetermined number of nine-node cells … Each cell has four integration points.” “The number of cells in the cross-sections influences the accuracy of section properties..”

Several methods are presented in those references, I assume that they are saying that the method used to determine k is that described in Ref. 1 by comparing strain energies. The numerical procedure in Ref. 1 is used and the cross section is divided into a number of Lagrangean nine node elements for numerical integration. However, the examples in Ref. 1 and Ref. 3 show that the shear factor depends on the Poisson ratio. ANSYS calculates the same factor for all values of the Poisson ratio (Table 2.1).

Value of N and T k for a Circle k for a Rectangle h/b=2 default 0.856914 0.842105 8 0.856293 0.833425 16 0.857088 0.833337 32 0.857139 0.833333 Exact for 0ν = 0.857143 0.833333

Table 2.1 Shear factor

The parameters N and T are the number of elements in each direction with a default value of N=T=4. See the SECDATA command. The table suggests that N=T=32 or greater will give accurate results for 0ν = . ANSYS uses it for all materials without correction for the actual Poisson ratio. 2.2. ANSYS solution for the cantilever beam. (a) End load only. The solution by ANSYS APDL using BEAM 188 and using one element with a square cross section, 2, 2, 4, 4 / 3b h A I= = = = , with 20L = , 1000E = , 0.3ν = , and using 0.833333k = , is

K3 = Linear quadratic cubic

3

3 ( )EIv LPL

= 0.80655 1.0078 1.0078

Table 2.2 End loaded cantilever

The quadratic and cubic models agree exactly with the analytical solution (1.17). The moment in this case varies linearly. If 100 elements are used, the solution for K3=0 also agrees with the exact value.

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(b) Cantilever with a uniformly distributed of intensity p (constant). For the case 2h b= = ,1, 1000, 0.3, 4 / 3, 4, 20, 0.833333,p E I A L kν= = = = = = = the solution using ANSYS APDL

and BEAM 188 is as follows:

Table 2.3 Distributed load

ANSYS does not allow pressure on lines for beam elements, you must use the command SFBEAM. The quadratic and cubic models agree exactly with the analytical solution (1.19). The moment in this case varies quadratically. If 100 elements are used, the solution for K3=0 agrees with the exact value. One Element is satisfactory for frame structures except for the linear interpolation (K3 = 0). Unfortunately, K3 = 0 is the default in APDL, so we need to set K3 = 2 or 3 for each problem. Workbench by default uses BEAM188 with K3 = 2. 2.3. Element Stiffness Matrix. Using one element and finding the reactions for a unit value of 1v , we can find the value of the first column of the stiffness matrix. For 20, 1000, .833333,L E k= = = and a square cross section with 2, 2,h b= = the first row of the stiffness matrix for K3 quadric or cubic is as follows. For 0,ν =

[ ] [ ]

1 1 2 2

1 1.9531 19.531 1.9531 19.531i

FY M FY Mk = −

(2.1)

For 0.3,ν =

[ ] [ ]

1 1 2 2

1 1.9395 19.395 1.9395 19.395i

FY M FY Mk = −

(2.2)

This agrees with the analytical result (1.11) in both cases. That is, the stiffness generated by ANSYS is identical to the stiffness matrix (1.11) derived directly from the differential equations of the Timoshenko model without internal nodes or numerical integration if the same value for k is used.

1 element 100 elements

K3 23

8EI vPL

23

8EI vPL

0 1.0754 1.0104 2 1.0104 1.0104 3 1.0104 1.0104

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REFERENCES 1. Schramm, U., L. Kitis, W. Kang, and W.D. Pilkey. “On the Shear Deformation Coefficient in Beam Theory.” [Finite Elements in Analysis and Design, The International Journal of Applied Finite Elements and Computer Aided Engineering]. 16 (1994): 141-162. 2. Pilkey, Walter D. [Formulas for Stress, Strain, and Structural Matrices]. Wiley, 1994. 3. Pilkey, Walter D. [Analysis and Design 0f Elastic Beams, Computational Methods]. Wiley, 2002. 4. Timoshenko, S., and J. N. Goodier. Theory of Elasticity. McGraw-Hill, 1951. 5. Dill, E. H. The Finite Element Method for Mechanics of Solids with ANSYS Applications. CRC Press. 2012.